Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
Follow up:Did you use extra space?
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?
空间复杂度要求为O(1),肯定需要考虑如何使用数组本身,使用数组的第一行和第一列作为标记。public void setZeroes(int[][] matrix) { if (!(matrix == null || matrix.length == 0 || matrix[0].length == 0)) { int lenI = matrix.length; int lenJ = matrix[0].length; boolean rowFirst = false; boolean colFirst = false; for (int i = 0; i < lenI; i++) { if (matrix[i][0] == 0) { colFirst = true; break; } } for (int j = 0; j < lenJ; j++) { if (matrix[0][j] == 0) { rowFirst = true; break; } } for (int i = 0; i < lenI; i++) { for (int j = 0; j < lenJ; j++) { if (matrix[i][j] == 0) { matrix[i][0] = 0; matrix[0][j] = 0; } } } for (int i = 1; i < lenI; i++) { if (matrix[i][0] == 0) { for (int j = 1; j < lenJ; j++) { matrix[i][j] = 0; } } } for (int j = 1; j < lenJ; j++) { if (matrix[0][j] == 0) { for (int i = 1; i < lenI; i++) { matrix[i][j] = 0; } } } if (rowFirst) { for (int j = 0; j < lenJ; j++) { matrix[0][j] = 0; } } if (colFirst) { for (int i = 0; i < lenI; i++) { matrix[i][0] = 0; } } } }
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原文地址:http://blog.csdn.net/u010378705/article/details/35568065